Lessons 42–43 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
43
The surface area of a sphere
Textbook, Part 2: pp. 165–170
GoalLearn to find the surface area of a sphere and of its parts and to use the relation V = ⅓ S R.
New words
surface of a sphere · sfera sirtispherical cap · sferik segmentspherical zone · sferik kamarsimilarity ratio · o‘xshashlik koeffitsiyenti
Explanation
The area of a sphere is S = 4πR², that is, four great circles. One way to see it: for a polyhedron circumscribed about the sphere V = ⅓ S′R (each face is the base of a pyramid with apex at the centre and height R); as the faces get small S′ → S and the polyhedron’s volume approaches that of the ball, so (4/3)πR³ = ⅓ S R and S = 4πR². The area of a spherical cap of height h is S = 2πRh; it depends only on the height, not on where the cap is placed. So a spherical zone of height h also has area 2πRh. For similar balls with ratio k of radii, the surfaces are in ratio k² and the volumes k³. If the radius triples, the surface grows 9 times and the volume 27 times. Recasting a ball into several smaller balls keeps the volume but increases the total surface.
Worked examples
R = 7: S = 4π · 49 = 196π; V = (4/3)π · 343 = 1372π/3. Check: ⅓ S R = ⅓ · 196π · 7 = 1372π/3. If S = 100π then 4R² = 100 and R = 5.
The cap of height h = 4 on a sphere of R = 10 has area 2π · 10 · 4 = 80π. A zone of a sphere with R = 5 between two parallel planes with height 3 has area 2π · 5 · 3 = 30π, wherever on the sphere the zone lies.
Class activity
“Archimedes’ peel”: with an adult’s help peel an orange and cut the peel into pieces. On paper draw four circles with the orange’s radius and cover them with the peel pieces — do they roughly fit? This demonstrates S = 4πR².
Practice
1
Find the surface of a sphere with R = 3.
36π
2
A sphere has surface 144π. Find its radius.
6
3
By what factor must the radius of a sphere be increased for its surface to grow 4 times?
2
4
Why is the area of a spherical zone 2πRh, independent of its position?
The zone’s area is proportional to the distance h between the planes and to the radius R: if we cut a sphere into layers of equal height all the layers have equal area (Archimedes: each equals the corresponding part of the circumscribed cylinder’s lateral surface, 2πR · h).